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The half-life of `.^215At` is `100mus`. The time taken for the activity of a sample of `.^215At` to decay to `1/16th` of its initial value isA. `400 mu s`B. `6.3 mu s`C. `40 mu s`D. `300 mu s` |
Answer» Correct Answer - a `(1)/(16)=(1)/(2^(t/100))` or `(1)/(2^(4))=(1)/(2^(t//100)) =4 =(t)/(100)` or `t=400 m u s`. |
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